1/(1x3)+1/(3x5)+1/(5x7)+1/(7x9)+1/(9x11).1/(97x99)=多少?这个肯定得用

ww120106kk2022-10-04 11:39:541条回答

1/(1x3)+1/(3x5)+1/(5x7)+1/(7x9)+1/(9x11).1/(97x99)=多少?这个肯定得用简便方法,请指教.

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caihuameimei 共回答了22个问题 | 采纳率81.8%: 您好
1/(1x3) = (1-1/3)/2
那么1/(1x3)+1/(3x5)+1/(5x7)+1/(7x9)+1/(9x11).1/(97x99)
=（1-1/3)/2+（1/3-1/5)/2+(1/5-1/7)/2+.+(1/97-1/99)/2
=(1-1/3+1/3-1/5+1/5-1/7+.-1/97+1/97-1/99)/2
=(1-1/99)X2
=49/99; 1年前

用数学归纳法证明1/(1x3)+1/(3x5)+1/(5x7)…1/(2n-1)(2n+1)=n/(2n+1) 用数学归纳法证明1/(1x3)+1/(3x5)+1/(5x7)…1/(2n-1)(2n+1)=n/(2n+1) 我证明完n=k+1后与结论不符,不知哪错了 当n=k时成立 即1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)=k/(2k+1) 则n=k+1时 1/(1x3)+1/(3x5)+1/(5x7)…1/(2k-1)(2k+1)+1/(2k+1)(2k+3) =k/(2k+1)+1/(2k+1)(2k+3) =2k^2+3k+1/(2k+1)(2k+3) =(k+1/2)(k+1)/(2k+1)(2k+3) =(k+1)/2(2k+3) 而原式应为(k+1)/(2k+3) zl88081年前1: a江心a 共回答了25个问题 | 采纳率96%

=k/(2k+1)+1/(2k+1)(2k+3)
=（2k+1）（k+1/（2k+3））
=（2k+1）（(2k方+3k+1)/（2k+3））
=1/（2k+1）*（(2k+1)（k+1）/（2k+3））
=（k+1）/（2k+3）成立
你算的
=(2k^2+3k+1）/(2k+1)(2k+3)
=（2k+1）（k+1）/(2k+1)(2k+3)
=（k+1）/（2k+3）
哪来的1/2

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1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007=? 1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007=? 例子：2/(1X3)=1-(1/3) 2/(3X5)=(1/3)-(1/5) 2/(5X7)=(1/5)-(1/7) 落英无痕361年前6: 贴完所有mm 共回答了15个问题 | 采纳率93.3%

你好：
1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007
=1/2×（1-1/3+1/3-1/5+1/5-1/7+.+1/2005-1/2007）
=1/2×（1-1/2007）
=1/2×2006/2007
=1003/2007

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1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007=? 1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007=? 例子：2/(1X3)=1-(1/3) 2/(3X5)=(1/3)-(1/5) 2/(5X7)=(1/5)-(1/7) 娥魅1年前5: xty91575 共回答了13个问题 | 采纳率84.6%

1/(1x3)+1/(3x5)+1/(5x7)+.+1/2005x2007
=（1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007）÷2
=（1-1/2007）÷2
=2006/2007÷2
=1003/2007

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